In a Cartesian system, I've got the slope, start point and distance of a line segment. What's the formula to find the endpoint?
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An equivalent way to Arturo's answer is as follows: from the slope $m$, you can determine the cosine and the sine of the angle from the horizontal axis of a line with that slope: $$c=\frac{1}{\sqrt{1+m^2}} \qquad s=\frac{m}{\sqrt{1+m^2}}$$ (exercise: verify that they are the cosine and sine of a certain angle) From this construction, you can easily determine the two points at a distance r from your starting point $(h,k)$ as $(h,k)\pm r(c,s)$. |
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If the point is $(a,b)$, then the distance from $(a,b)$ to $(x,y)$ is $$\sqrt{(x-a)^2 + (y-b)^2}.$$ If the point is $(a,b)$, then the points that lie on the line through $(a,b)$ with slope $m$ are the points of the form $$(a,b) + k(1,m)$$ where $k$ is a constant. Putting the two together, if you know the start point $(a,b)$, and the slope $m$, and the distance $d$, then find the (two) values of $k$ that will give you a distance of $d$ by plugging in and solving for $k$. |
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Y=mx+c is the equation of the line you have. (x1,y1) is the point and D is the distance. (x,y) is the point you don't know. D= sqrt((x1-x)^2 +(y1-y)^2) sub in for y D= sqrt((x1-x)^2 +(y1-(mx+c))^2) then solve for the only unknown, x. this is your x co-ord (2 values). then y=mx+c gives the y. |
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